Integer division that rounds up

Introduction

Usual integer division rounds down: ${\frac {a}{b}}=\left\lfloor {\frac {a}{b}}\right\rfloor$ for $a,b\in \mathbb {N} ,b\neq 0$. To round up (if overflow is not an issue), you can use following algorithm with the usual roundig down division: $q={\frac {x+y-1}{y}}=\left\lceil {\frac {x}{y}}\right\rceil$

Proof

Proof is in two parts; 1st if $y$ divides $x$, and if not. Note that usual integer division rounds down.

Part 1. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} divides Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=ay} for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\in\mathbb N_+} . Thus we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \left \lfloor \frac{x+y-1}{y} \right \rfloor &= \left \lfloor \frac{x}{y} + \frac{y-1}y \right \rfloor \\ &= \left \lfloor \frac{ay}{y} + \frac{y-1}y \right \rfloor \\ &= \frac{ay}{y} \\ &= \frac{x}{y} \end{align} }

because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 \leq \frac{y-1}y < 1} . This part is ok.

Part 2. If $y$ does not divide $x$ we have $x=by+r$ for some $b\in \mathbb {N} _{+}$ and $0<r<b$. Thus we have

${\begin{aligned}\left\lfloor {\frac {x+y-1}{y}}\right\rfloor &=\left\lfloor {\frac {x}{y}}+{\frac {y-1}{y}}\right\rfloor \\&=\left\lfloor {\frac {by+r}{y}}+{\frac {y-1}{y}}\right\rfloor \\&=\left\lfloor {\frac {by}{y}}+{\frac {r}{y}}+{\frac {y}{y}}-{\frac {1}{y}}\right\rfloor \\&=\left\lfloor {\frac {by+y}{y}}+{\frac {r-1}{y}}\right\rfloor \\&=\left\lfloor {\frac {(b+1)y}{y}}+{\frac {r-1}{y}}\right\rfloor \\&={\frac {(b+1)y}{y}}\\\end{aligned}}$

Thus we have $\left\lfloor {\frac {x+y-1}{y}}\right\rfloor =\left\lceil {\frac {x}{y}}\right\rceil$